Question

A block of mass 4 Kg is acted upon by a 50 N force as shown. The friction coefficient between block and wall is $\mu$

• A. $For\mu =0.5$block will be at rest
• B. $For\mu$ = 0.2 block will move down
• C. $For\mu$ = 0.8 block will move up
• D. Block can never move up for any value of $\mu$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $For\mu =0.5$block will be at rest
B $For\mu$ = 0.2 block will move down
D Block can never move up for any value of $\mu$

Resolving the applied force into vertical and horizontal components, we obtain the given FBD.

It is clear from the FBD $N=50\cos {37}^{\circ }=40N$

In the vertical direction, net external force is: $F_{net}=mg-50\sin {37}^{\circ }=10N$ downwards.

Thus, since net external force is downwards it is clear that friction will always act upwards.

To keep the block at rest, $f_{max}\ge 10N$, otherwise, it will move down. We know that $f_{max}=\mu N=40\mu$. Thus:

$40\mu \ge 10\Rightarrow \mu \ge 0.25$ to prevent downward sliding.

$\therefore$ options $\text{(A), (B), (D)}$ are correct.

$\to QED.$